Are you trying to define the Fourier transform for $\mathbb{F}_7$, or are you trying to trying to find the Fourier transform of a multiplicative character on $\mathbb{F}_7^*$? My guess is the latter based on the fact that there are $q-2$ nontrivial multiplicative characters on $\mathbb{F}_q$.
–
JavaManApr 22 '11 at 19:07

2

I would edit it if I could, but technically the term "multiplicative character on $\mathbb{F}_7^*$" is redundant. It should read either multiplicative character of $\mathbb{F}_7$ or character of $\mathbb{F}_7^*$.
–
JavaManApr 22 '11 at 19:29

1 Answer
1

Let $\mathbb{F}_q^d$ denote the $d$-dimensional vector space over the finite field with $q$ elements. Recall that for a function $f : \mathbb{F}_q^d \to \mathbb{C}$, the Fourier transform of $f$ is given by

where $\chi$ is a nontrivial additive character on $\mathbb{F}_q$. It is worth noting that this definition of $\widehat{f}$ may or may not agree with the definition you use. In particular, your definition might not include the normalization factor $q^{-d}$.

Now, when $q$ is prime, we can actually take $\chi(z) = \exp(2 \pi i z/q)$. Furthermore, when $q$ is prime, we can identify $\mathbb{F}_q$ with $\mathbb{Z}_q = \{0,1, \dots , q-1\}$.